Ecoop2024_TIforWildFJ/conclusion.tex

\section{Properties of the Algorithm}

%TODO: how are the challenges solved: Describe this in the last chapter with examples!
\section{Soundness}\label{sec:soundness}

\section{Completeness}\label{sec:completeness}
The algorithm can find a solution to every program which the Unify by Plümicke finds
a correct solution aswell.
It will not infer intermediat type like $\wctype{\rwildcard{X}}{Pair}{\rwildcard{X},\rwildcard{X}}$.
There is propably some loss of completeness when capture constraints get deleted.
This happens because constraints of the form $\tv{a} \lessdotCC \exptype{C}{\wtv{x}}$ are kept as long as possible.
Here the variable $\tv{a}$ maybe could hold a wildcard type,
but it gets resolved to a $\generics{\type{A} \triangleleft \type{N}}$.
This combined with a constraint $\type{N} \lessdot \wtv{x}$ looses a possible solution.

This is our result:
\begin{verbatim}
class List<X> {
    <Y extends X> void addTo(List<Y> l){
        l.add(this.get());
    }
}
\end{verbatim}
$
\tv{l} \lessdotCC \exptype{List}{\wtv{y}},
\type{X} \lessdot \wtv{y}
$
But the most general type would be:
\begin{verbatim}
class List<X> {
    void addTo(List<? super X> l){
        l.add(this.get());
    }
}
\end{verbatim}

\subsection{Discussion Pair Example}
\begin{verbatim}
<X> Pair<X,X> make(List<X> l){ ... }
<X> boolean compare(Pair<X,X> p) { ... }

List<?> l;
Pair<?,?> p;

compare(make(l)); // Valid
compare(p); // Error
\end{verbatim}

Our type inference algorithm is not able to solve this example.
When we convert this to \TamedFJ{} and generate constraints we end up with:
\begin{lstlisting}[style=tamedfj]
let m = let x = l in make(x) in compare(m)
\end{lstlisting}
\begin{constraintset}$
\wctype{\rwildcard{X}}{List}{\rwildcard{X}} \lessdot \ntv{x},
\ntv{x} \lessdotCC \exptype{List}{\wtv{a}}
\exptype{Pair}{\wtv{a}, \wtv{a}} \lessdot \ntv{m}, %% TODO: Mark this constraint
\ntv{m} \lessdotCC \exptype{Pair}{\wtv{b}, \wtv{b}}
$\end{constraintset}

$\ntv{x}$ will get the type $\wctype{\rwildcard{X}}{List}{\rwildcard{X}}$ and
from the constraint
$\wctype{\rwildcard{X}}{List}{\rwildcard{X}} \lessdot \exptype{List}{\wtv{a}}$
\unify{} deducts $\wtv{a} \doteq \rwildcard{X}$ leading to 
$\exptype{Pair}{\rwildcard{X}, \rwildcard{X}} \lessdot \ntv{m}$.

Finding a supertype to $\exptype{Pair}{\rwildcard{X}, \rwildcard{X}}$ is the crucial part.
The correct substition for $\ntv{m}$ would be $\wctype{\rwildcard{X}}{Pair}{\rwildcard{X}, \rwildcard{X}}$.
But this leads to additional branching inside the \unify{} algorithm and increases runtime.
%We refrain from using that type, because it is not denotable with Java syntax.
%Types used for normal type placeholders should be expressable Java types. % They are not!

The prefered way of dealing with this example in our opinion would be the addition of a multi-let statement to the syntax.

\begin{lstlisting}[style=letfj]
let x : (*@$\wctype{\rwildcard{X}}{List}{\rwildcard{X}}$@*) = l, m = make(x) in compare(make(x))
\end{lstlisting}

We can make it work with a special rule in the \unify{}.
But this will only help in this specific example and not generally solve the issue.
A type $\exptype{Pair}{\rwildcard{X}, \rwildcard{X}}$ has atleast two immediate supertypes:
$\wctype{\rwildcard{X}}{Pair}{\rwildcard{X}, \rwildcard{X}}$ and
$\wctype{\rwildcard{X}, \rwildcard{Y}}{Pair}{\rwildcard{X}, \rwildcard{Y}}$.
Imagine a type $\exptype{Triple}{\rwildcard{X},\rwildcard{X},\rwildcard{X}}$ already has 
% TODO: how many supertypes are there?
%X.Triple<X,X,X> <: X,Y.Triple<X,Y,X> <: 
%X,Y,Z.Triple<X,Y,Z>

% TODO example:
\begin{lstlisting}[style=java]
<X> Triple<X,X,X> sameL(List<X> l)
<X,Y> Triple<X,Y,Y> sameP(Pair<X,Y> l)
<X,Y> void triple(Triple<X,Y,Y> tr){}

Pair<?,?> p ...
List<?> l ...

make(t) { return t; }
triple(make(sameP(p)));
triple(make(sameL(l)));
\end{lstlisting}

\begin{constraintset}
$
\exptype{Triple}{\rwildcard{X}, \rwildcard{X}, \rwildcard{X}} \lessdot \ntv{t},
\ntv{t} \lessdotCC \exptype{Triple}{\wtv{a}, \wtv{b}, \wtv{b}}, \\
(\textit{This constraint is added later: } \exptype{Triple}{\rwildcard{X}, \rwildcard{Y}, \rwildcard{Y}} \lessdot \ntv{t})
$
% Triple<X,X,X> <. t
% ( Triple<X,Y,Y> <. t ) <- this constraint is added later
% t <. Triple<a?, b?, b?>

% t =. x.Triple<X,X,X>
\end{constraintset}


\section{Conclusion and Further Work}
% we solved the problems given in the introduction (see examples TODO give names to examples)
The problems introduced in the openening \ref{challenges} can be solved via our \unify{} algorithm (see examples \ref{example1} and \ref{example2}).
As you can see by the given examples our type inference algorithm can calculate
type solutions for programs involving wildcards.

Going further we try to proof soundness and completeness for \unify{}.


% The tricks are:
% \begin{itemize}
% \item Erasing wildcards by setting $\type{U} \doteq \type{L}$
% \end{itemize}

% Improvements:
% \begin{itemize}
% \item Straightforward algorithm. Less data types used. Only two sort of constraints.
% \end{itemize}

% \subsection{Problems}
% The algorithm is lacking completeness.

% \begin{verbatim}
% <A> void m(List<A> a){}

% m2(l){
%     l.add("String");
% }
% \end{verbatim}

%COnstraints:
%l <. List<a>
%String <. a